Question

prove Heron's formula for area of a triangle whose sides are given a,b,c

Answer 1

prove Heron's formula for area of a triangle whose sides are given a,b,c

The demonstration and proof of Heron's formula can be done from elementary consideration of geometry and algebra. I will assume the Pythagorean theorem and the area formula for a triangle

$$A = \frac {1}{2} b h,$$

where $b$ is the length of a base and $h$ is the height to that base.

Let $a$, $b$, and $c$ be the lengths of the sides of our triangle and $h$ the height to the side of length $c$.

We have

$$S = \frac {a + b + c}{2},$$

so, for future reference,

$$\begin{array}{l} 2 s = a + b + c \\ 2 (s - a) = - a + b + c \\ 2 (s - b) = a - b + c \\ 2 (s - c) = a + b - c \\ \end{array}$$

There is at least one side of our triangle for which the altitude lies "inside" the triangle. For convenience make that the side of length $c$. It will not make any difference, just simpler.

Our task is to express $h$ in terms of $a$, $b$, and $c$, then substitute for $h$ in $A = \frac{1}{2} ch$.

Let $p + q = c$ as indicated. Then

$$h^2 + p^2 = a^2$$

and

$$h^2 + q^2 = b.$$

Since

$$q = c - p,$$

then

$$q^2 = (c - p)^2$$

and

$$q^2 = c^2 - 2cp + p^2.$$

Adding $h^2$ to each side gives

$$h^2 + q^2 = h^2 + c^2 - 2cp + p^2.$$

Now substitute in this equation and get $b^2 = a^2 - 2cp + c^2$ and solve for $p$ to get

$$p = \frac{a^2 + c^2 - b^2}{2c}.$$

Now, since $h^2 = a^2 - p^2$ we can substitute for $p$ and get an expression in terms of $a$, $b$, and $c$.

$$h^2 = a^2 - p^2 =$$ $$(a + p)(a - p) =$$ $$\left(a + \frac{a^2 + c^2 - b^2}{2c}\right)\left(a - \frac{a^2 + c^2 - b^2}{2c}\right) =$$ $$\left(\frac{2ac + a^2 + c^2 - b^2}{2c}\right)\left(\frac{2ac - a^2 - c^2 + b^2}{2c}\right) =$$ $$\frac{((a + c)^2 - b^2)(b^2 - (a - c)^2)}{4c^2} = \frac{(a + b + c)(-a + b + c)(a + b + c)(a + b - c)}{4c^2} =$$ $$\frac {2 s * 2 (s - a) * 2 (s - b) * 2 (s - c)}{4 c ^ {2}}$$

Therefore

$$h ^ {2} = \frac {4 s (s - a) (s - b) (s - c)}{c ^ {2}}$$ $$h ^ {2} = \frac {2 \sqrt {s (s - a) (s - b) (s - c)}}{c}$$

Since

$$A = \frac {1}{2} c h,$$

then

$$A = \frac {1}{2} c \frac {2 \sqrt {s (s - a) (s - b) (s - c)}}{c},$$

and

$$A = \sqrt {s (s - a) (s - b) (s - c)}.$$